# Fourier Transform of function with measurement uncertainty

I'm attempting to create a classifier that will classify two different types of functions, the problem is, typically the functions have significant noise. (This noise is from a dark current, and is therefore actually Poisson in nature; however, in the simulated data I used Gaussian distributed white noise for simplicity, and it's a good approximation anyway.)

### Simulated Experimental $$f'(x) = f(x) + \mathcal{N}(0,\sigma)$$ and $$g'(x) = g(x) + \mathcal{N}(0,\sigma)$$:

We can take a DFT of both waveforms, and they match up reasonably well.

### DFT:

However, as you can see, there definitely error that can confound classification. So to help analyze the situation, I'd like to be able to put error bars on the Fourier transformation itself.

How does one approach taking a Fourier transform of a function with an associated error?

Unfortunately, most Google Searches on this topic talk about the Fourier Transform of the Error Function itself, or of the Gaussian, or about Heisenberg's Uncertainty Principle, depending on the search terms used, which are not what I'm looking for.

Apologizes in advance for rampant and unrepentant abuse of notation.

A real valued Normal distribution with a mean of 0 and a standard deviation of $\sigma$ has a Fourier transform of that's simply:

$$\hat{\mathcal{N}}(f, \mu, \sigma) = \begin{cases} \mu \delta(0) + \mathcal{N}(0, \sigma) & f = 0 \\ \mathcal{CN}(0, \sigma) & f \ne 0 \\ \end{cases}$$

Where $\mathcal{CN}(\mu,\sigma)$ is a circularly symmetric complex-valued Normal Distribution about the complex mean $\mu$ and the real-valued standard deviation $\sigma$.

This is because for $f = 0$, we essentially average the probability sum of normal distributions, which just gives back a normal distribution. We can use $\mathcal{N}(\mu, \sigma) = \mu + \mathcal{N}(0, \sigma)$ and exploit the linearity of the Fourier transform.

For $f \ne 0$ then the normal distribution is smeared around the complex plane by the Fourier transform. Consider a point $z = x + iy$ that is a distance $r$ from the mean, which is taken to be 0 without loss of generality (because, again, linearity).

What is the probability density of $z$?

$$\begin{split} z &= \mathcal{P}\left(\hat{\mathcal{N}}(f, \mu, \sigma)\right) \\ z &= \mathcal{P}\left(\int_{-\infty}^{\infty} \mathcal{N}(t,\mu,\sigma) \mathrm{e}^{-2 \pi f t} dt\right) \\ \end{split}$$

The normal distribution is time invariant and $f \ne 0$

$$z = \mathcal{P}\left(\int_{-\infty}^{\infty} \mathcal{N}(0,\sigma) \mathrm{e}^{-2 \pi f t} dt\right)$$

Next, we can swap the order of the integral and the probability function, this is because each probability at time t is independent, so the sum of the probabilities is the same as the probability of the sum.

$$z = \int_{-\infty}^{\infty}\mathcal{P}\left( \mathcal{N}(0,\sigma) \mathrm{e}^{-2 \pi f t} \right)dt$$

$z$ only can get contributions from cases where the arg of the differential slice is equal to either $\mathrm{Arg}(z)$ or $\mathrm{Arg}(-z)$. So these values are depressed by $\frac1{\pi}$ compared to the regular normal distribution at a distance $r$ from the mean, which is what we would expect if no other change to the distribution has occurred other than smearing around the complex plane. This is a little hand-wavey, but I have no idea how to notate this properly.

However, I am taking the discrete fourier transform, which privileges certain frequencies that are certain simple multiples of the sampling frequency. In these cases the distribution may not be circular. The good news is for my application, the details in this particular domain is not important, because my signals do not have significant power in these frequencies (which tend to be much higher.)